All categories

3 years ago

A bird is flying north with a mass of 2.5kg has a momentum of 17.5kg m/s at what velocity is it flying

Physics

2 answers:

klasskru [66]3 years ago

8 0

Momentum = mass • velocity
v= 17.5/2.5
= 7 m/s

Law Incorporation [45]3 years ago

7 0

Answer:

v=7 m/s

Explanation:

momentum definition

The amount of movement or linear momentum is the product of velocity by mass. Velocity is a vector while mass is a scalar. As a result we obtain a vector with the same direction and direction as the velocity.

The magnitude of the momentum is calculated like this:

p=m*v Formula (1)

where:

p: momentum (kg*m/s)

m: mass (kg)

v: velocity (m/s)

Data

m = 2.5kg

p = 17.5kg m/s

Problem development

We replace the data in the formula (1)

17.5kg m/s = 2.5kg*v

$17.5 kg*\frac{m}{s} = 2.5kg * v$

$v=\frac{17.5 kg*\frac{m}{s} }{2.5 kg}$

v = 7 m/s

You might be interested in

Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri

irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

$d=\frac{2g(m1-m2)}{k}$

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

$E_{i}=E_{f}$

$K_{i}+U_{i}=K_{f}+U_{f}$ (1)

With $K_{i}$ the initial kinetic energy, $U_{i}$ the initial potential energy, $K_{f}$ the final kinetic energy and $U_{f}$ the final potential energy. Note that initialy the masses are at rest so $K_{i} = 0$ , when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so $K_{f} =0$ . So, equation (1) becomes:

$U_{i}=U_{f}$ (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so $U_{i}=U_{gi}$ and at final moment we have potential gravitational energy and potential elastic energy so $U_{f}=U_{gf}+U_{ef}$ , using this on (2)

$U_{gi}=U_{gf}+U_{ef}$ (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), $U_{gi}=0$ and we have by (3) :

$0= U_{gf}+U_{ef}$ (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

$0=-m1gd+m2gd+\frac{kd^{2}}{2}$

dividing both sides by d

$0=-m1g+m2g+\frac{kd}{2}$

$g(m1-m2)= \frac{kd}{2}$

$d=\frac{2g(m1-m2)}{k}$ , with k the constant of the spring and g the gravitational acceleration.

7 0

3 years ago

How are the forces observed related to the motion of the earth around the sun

Inga [223]

This question sounds like it came after some activity where
some forces were observed. Since we were not there, and
we don't know what the activity was, we don't know what forces
were observed, and we have no clue to how they might be related
to the motion of the Earth around the sun.

7 0

3 years ago

Explain why it is not advisable for soldiers to march across the bridge in rythm

AnnyKZ [126]

Answer:

might collapse

Explanation:

y'all r too heavy and tht shi will collapse and all y'all would die

4 0

1 year ago

Read 2 more answers

6. Draw conclusions: Newton’s first law states that an object in motion will travel at a constant velocity unless acted upon by

gavmur [86]

You can test if it’s true by holding a pencil in mid air over a table and the table is supposed to be the unbalanced forced that stopped the pencil from moving at the constant velocity it was going by.

5 0

3 years ago

Read 2 more answers

A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient

gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0

2 years ago